Find the integer $n,$ $-90 < n < 90,$ such that $\tan n^\circ = \tan 1000^\circ.$
Explanation: Since the tangent function has period $180^\circ,$
\[\tan 1000^\circ = \tan (1000^\circ - 6 \cdot 180^\circ) = \tan (-80^\circ),\]so $n = \boxed{-80}.$